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Remainder After Division Problem

  1. Jul 27, 2013 #1
    1. The problem statement, all variables and given/known data
    (-8)^4124 + 6^3101 + 7^5 is divided by 3.


    2. Relevant equations



    3. The attempt at a solution
    My original insight was that 6 raised to any power is always divisible by 3.
    7 raised to any power yields a remainder of 1 when divided by 3.
    and the remainder from -8 raised to a power alters from -2 (when exponent is odd) to 1 (when exponent is even.

    So in my problem the first term leaves a remainder 1, the 2nd term no remainder and the final term remainder 1. What I am confused by, is how is this all effected when we add the numbers? My immediate intuition would tell me to add the remainders but I am sure that is incorrect.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 27, 2013 #2

    ehild

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    Why do you think it is incorrect? :smile:


    ehild
     
  4. Jul 27, 2013 #3

    pasmith

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    You need to calculate ((-8)^4124 + 6^3101 + 7^5) mod 3.

    You've already found that 6 mod 3 = 0 and 7 mod 3 = 1, but you haven't found -8 mod 3 correctly. Remember: you are trying to write -8 as 3n + k, where n is an integer (possibly negative) and k = 0, 1 or 2.
     
  5. Jul 27, 2013 #4
    Thanks for the hint! I am still a little bit confused, according to Wolfram the remainder is 2 indeed the sum of the remainder from each of the divisions separately..


    -8^4124 /3 --> r= 1
    6^3101 /3 --> r= 0
    7^5 /3 --> r =1

    1+0+1 =2

    Was this a fluke or is this actually the consistent method?
     
  6. Jul 28, 2013 #5

    ehild

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    -8=-9+1. (-8)n=(-9+1)n. According to the Binomial Theorem, any positive integer power of -8 contains terms divisible by 9 except the last one which is the power of 1.
    7=6+1. Again, any positive integer power 7n=(6+1)n contains terms divisible by 6, except the last one, which is 1.

    Adding up terms divisible by three, the sum is also divisible by three. And you have the two "1" terms to add.


    ehild
     
  7. Jul 29, 2013 #6
    Thanks ehild!
     
  8. Jul 30, 2013 #7
    Answer : 2
    First of all remove - sign from the first term as it is to the even power.
    You'll get (9-1)^4124 Now the only term which is independent of 9 in the binomial series expansion will be (-1)^1024 which gives remainder 1 .
    Second Term is divisible .Remainder 0
    Write third term as (6+1)^5 .From here again you'll get remainder 1.
    ADD THE Remainders : 1+0+1 = 2 (ANSWER)
     
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